# How do you integrate int x^3e^(x^2-1)dx using integration by parts?

Nov 9, 2015

See the explanation.

#### Explanation:

$\int {x}^{3} {e}^{{x}^{2} - 1} \mathrm{dx}$

Integrateing ${x}^{3}$ and differentiating ${e}^{{x}^{2} - 1}$ would give us MORE $x$'s, so let's try the other way around first.

To integrate ${e}^{{x}^{2} - 1}$, we'll need another $x$ so that we can substitute.

So,
$\int {x}^{3} {e}^{{x}^{2} - 1} \mathrm{dx} = \int {x}^{2} \left[x {e}^{{x}^{2} - 1} \mathrm{dx}\right]$

Let $u = {x}^{2}$ and $\mathrm{dv} = x {e}^{{x}^{2} - 1} \mathrm{dx}$.

We get $\mathrm{du} = 2 x \mathrm{dx}$ and $v = \frac{1}{2} {e}^{{x}^{2} - 1}$.

Our integral becomes

$= \frac{1}{2} {x}^{2} {e}^{{x}^{2} - 1} - \int x {e}^{{x}^{2} - 1} \mathrm{dx}$.

The integral may again be evaluated by substitution.

$= \frac{1}{2} {x}^{2} {e}^{{x}^{2} - 1} - \frac{1}{2} {e}^{{x}^{2} - 1} + C$.